Maximal filter and its topology in lattice implication algebras

Abstract

The present work aims at investigating maximal filter and its topological properties in lattice implication algebras (LIAs). To begin with, the maximal filters of LIAs are investigated and some equivalent characterizations of maximal filters are also given. In addition, the radical of filters in a LIA is introduced and its properties are also obtained, the structure of radical of maximal filter is investigated. Finally, by introducing some topological structures on the set of all maximal filters and by investigating the topological properties of them, we conclude that the set of all maximal filters is a compact topological space and the set of all maximal filters is also a Hausdorff topological space.

Keywords

Lattice implication algebra maximal filter radical topology

1 Introduction

Logic gives a technique for laying the foundations of an important task which makes to the computer simulate human being in dealing with certainty in information in the field of artificial intelligence (AI). While information processing dealing with certain information is based on the classical logics. However, there are many uncertainties in real world, such as fuzziness, randomness, imprecisions, etc. Classical logics are invalid to provide the logical foundation for the information with some uncertainties. Therefore, non-classical logic has become as a formal and useful tool for computer to deal with uncertain information. Many-valued logic, a great extension and development of classical logic, has always been a crucial direction in non-classical logic. Lattice-valued logic, a kind of very important many-valued logic, plays an important role for two aspects: (1) it extends the chain-type truth-valued field of some well known logic to some relatively general lattice; (2) the incompletely comparable property of truth value characterized by general lattice can more efficiently reflect the uncertainty of human being’s thinking, judging and decision. Hence, lattice-valued logic is becoming a research field which strongly influences the development of algebraic logic, computer science and artificial intelligent technology. In order to research the many-valued logical system whose truth value is given in a lattice, Xu [19] firstly established the lattice implication algebra (LIA) by combining lattice and implication algebra, and investigated many useful structures [3–8 , 25]. LIA provides the foundation to establish the corresponding logical system from the algebraic viewpoint. As it is known, filter theory plays vital role in proof theory of a logical systems and the completeness of corresponding semantic model. From the view of logic, filter corresponds to the set of provable formulas in the corresponding propositional logical systems, and different filter corresponds to different provable formulas set. Therefore, filter plays a vital role in various logical algebras and it is an important tool in researching logical algebras. Therefore, filter theories of LIAs are very important direction in the development of LIA and also attracted many scholar’s attention Since LIA’s introduction. For instance, Xu and Qin [22] introduced the notions of filter and implicative filter in a lattice implication algebra, and investigated their properties [20, 22]. Since then, many scholars studied various of filters in LIAs [3–8 , 22].

Studying the topological structure [17] is another important aspect on the research of lattice theories. Wang [18] investigated the topological characterizations of consistency theory in classical logic [18] and Zhou [28] gave the characterizations of maximal consistency theory in classical propositional logic and constructed the topology of the set of maximal consistency theory [29]. As it is known, filter theory plays vital role in proof theory of a logical systems and the completeness of corresponding semantical model, the topological properties of logical algebras are investigated from the point of topology by using the filter theory. These thoughts revealing the links between topological spaces and posets are constantly carried forward, and by the impact of them some new branch of mathematics such as continuous lattice theories, frame theories, etc, also resulting in [2]. In recent years, with the rapid development of fuzzy logic, some lattices with new properties, such as FI-algebras, lattice implication algebras, BCI-algebras, BL-algebras, MTL-algebras, etc, have also been proposed. Investigating the relationship between these new posets and topological spaces are naturally important research topic. As a matter of fact, some authors have already studied the topological structures for some logic algebras, such as MTL-algebra, triangle algebras and residuated lattices [1 , 23–27].

The main goal of this work is to introduce the topological method to lattice implication algebra and investigate the maximal filter of lattice implication algebras and its topological properties. The rest of the paper is organized as follows. In Section 2, we review some definitions and properties on lattice implication algebra, which are used in the analysis throughout this paper. In Section 3, we introduce the maximal filter of lattice implication and radical of lattice implication algebra, their properties and equivalent characterizations are studied. In Section 4, by introducing some topological structures on the set of all maximal filters and by investigating its topological properties, we conclude that the set of all maximal filters is a compact Hausdorff topological space. This paper is concluded in Section 5.

2 Preliminaries

In the following, we review some elementary concepts and conclusions of a lattice implication algebra. More details of LIAs can be found in [21].

Definition 2.1. ([21]) Let (L, ∨, ∧, O, I) be a bounded lattice with an order-reversing involution ′, the greatest element I and the smallest element O, and $\to : L \times L ⟶ L$ be a mapping. $L = (L, \lor, \land,^{'}, \to, O, I)$ is called a lattice implication algebra if the following conditions hold for any x, y, z ∈ L:

(I₁) x → (y → z) = y → (x → z);

(I₂) x → x = I;

(I₃);

(I₄) x → y = y → x = I implies x = y;

(I₅) (x → y) → y = (y → x) → x;

(I₁) (x ∨ y) → z = (x → z) ∧ (y → z);

(I₂) (x ∧ y) → z = (x → z) ∨ (y → z).

In this paper, we denote

L

as a lattice implication algebra (L, ∨, ∧, ′, →, O, I).

We list some basic properties of lattice implication algebras. It is useful to develop these topics in other sections.

Example 2.1. Let $L = {O, a, b, c, d, I}$ , the Hasse diagram of L be defined as Figure 1 and its implication operator → be defined as Table 1 and operator ′be defined as Table 2. Then $L = (L, \lor, \land,^{'},$ →, O, I) is a lattice implication algebra.

Fig. 1

Hasse Diagram of L.

Table 1

$\to$ of $L$

→	O	a	b	c	d	I
O	I	I	I	I	I	I
a	c	I	b	c	b	I
b	d	a	I	b	a	I
c	a	a	I	I	a	I
d	b	1	I	b	I	I
I	O	a	b	c	d	I

Table 2

′ of $L$

′
O	I
a	c
b	d
c	a
d	b
I	O

Example 2.2.(Lukasiewicz implication algebra on finite chain) Let L_n = {a_i|i = 1, 2, ⋯, n}, a₁ ≤ a₂ ≤ ⋯ ≤ a_n. For any 1 ≤ j, k ≤ n, define

\begin{array}{l} a_{j} \lor a_{k} = a_{m a x {j, k}}, a_{j} \land a_{k} = a_{m i n}_{{j, k}}, \\ (a_{j})^{'} = a_{n - j + 1}, a_{j} \to a_{k} = a_{m i n}_{{n - j + k, n}} . \end{array}

Then (L_n, ∨, ∧, ′, →, a₁, a_n) is a lattice implication algebra.

Theorem 2.1. ([21]) Let $L$ be a lattice implication algebra. Then for any x, y, z ∈ L, the following conclusions hold:

if I → x = I, then x = I;

I → x = x and x → O = x′;

O → x = I and x → I = I;

(x → y) → ((y → z) → (x → z)) = I;

(x → y) ∨ (y → x) = I;

if x ≤ y, then x → z ≥ y → z and z → x ≤ z → y;

x ≤ y if and only if x → y = I;

(z → x) → (z → y) = (x ∧ z) → y = (x → z) → (x → y);

x → (y ∨ z) = (y → z) → (x → z);

x → (y → z) = (x ∨ y) → z if and only if x → (y → z) = x → z = y → z;

z ≤ y → x if and only if y ≤ z → x.

Definition 2.2. ([16]) A non-empty subset F is a filter of a lattice implication algebra $L$ if and only if

(F3) I ∈ F.

(F4) if x ∈ F, x → y ∈ F, then y ∈ F.

In lattice implication algebras, define the binary operation ⊗, ⊕ as follows: for any x, y ∈ L,

x ⊗ y = (x → y′)′;

x ⊕ y = x′ → y.

Some properties of operation ⊗, ⊕ are listed as follows, also can refer to reference [21].

Theorem 2.2. [21] Let $L$ be a lattice implication algebra, for any x, y ∈ L, the following properties:

x ⊗ y = y ⊗ x, x ⊕ y = y ⊕ x;

(x ⊗ y)′ = x′ ⊗ y′, (x ⊕ y)′ = x′ ⊗ y′;

(O ⊗ x)′ = O, I ⊗ x = x, x, ⊗ x′ = O (x ⊕ y)′ = x′ ⊗ y′;

O ⊗ x = O,I ⊗ x = x = x, x = ⊗ x′ = O;

x → (x ⊗ y) = (x ⊕ y) → y;

x → (y → z) = (x ⊗ y) → z;

x ≤ y → z if and only if x ⊗ y ≤ z.

Proposition 2.2. ([6]) A non-empty subset F of a lattice implication algebra $L$ is a filter of $L$ if and only if it satisfies

(F1) if x, y ∈ F, then x ⊗ y ∈ F.

(F2) if x ∈ F and x ≤ y, then y ∈ F.

Definition 2.3. ([21]) Let $L$ be a lattice implication algebra, F ⊆ L is said to be is said to be an implicative filter of $L$ , if it satisfies the following conditions:

(F3) I ∈ F;

(F5) for any x, y, z ∈ L, if x → (y → z) ∈ F and x → y ∈ F, then x → z ∈ F.

Definition 2.4. ([21]) Let F be a filter of a lattice implication algebra $L$ . F is a prime filter if and only if

x ∨ y ∈ F implies x ∈ F or y ∈ F for any x, y ∈ L.

Theorem 2.3. ([16]) Let $L$ be a lattice implication algebra and F an implicative filter of L, then for any x, y ∈ L, (x → y) → x ∈ F implies x ∈ F.

Definition 2.5. ([16]) Suppose A ⊆ L, the least filter containing A is called the filter generated by A and denoted by [A). It is trivial to verify that

[A) = ∩ {B|A ⊆ B ⊆ L, B is a filter}.

In what follows, [{a}) is denoted by [a).

Theorem 2.4. ([21]) Let $L$ be a lattice implication algebra, ∅ ≠ A ⊆ L. Then

[A) = {x|x ∈ L, there exist a₁, ⋯, a_n ∈ A, such that a₁ ⊗ ⋯ ⊗ a_n ≤ x}.

Obviously, [L) = L.

3 Maximal filters and its radical in lattice implication algebras

In this section, we introduce the notion of maximal filter in a lattice implication algebra, and properties of maximal filters will be discussed in this section.

3.1 Maximal filters

Definition 3.1. Let $L$ be a lattice implication algebra and F be a proper filter of $L$ . F is said to be a maximal filter of $L$ if and only if it is not included in any other proper filter of $L$ .

Remark 1. Let F be a filter of $L$ . F is a maximal filter of $L$ if and only if F ⊆ J implies J = F or J = L for any filter J of $L$ .

Example 3.1. In Example 2.1, let F₁ = {I, a}, F₂ = {I, b, c}, it is routine to verify F₁ and F₂ are maximal filters of $L$ .

Theorem 3.1. Let F ba a filter of $L$ that satisfies:

(M1) x ∈ F if and only if for any x ∈ L.

Then F is a maximal filter of $L$ .

Proof. Suppose that F is a filter of $L$ satisfies that (M1), as I ∈ F, then O ∉ F, therefore F is proper filter. If F ⊆ J and J is also a proper filter. According to Definition 3.1, we need to prove J = F. In fact,if there is x ∈ J such that x ∉ F. It follows that, that is, x → O ∈ F ⊆ J. And so O ∈ J and J = L, which is a contradiction.

Definition 3.2. Let $L$ be a lattice implication. A subset F of $L$ is said to have the finite intersection property if $a_{1} \otimes a_{2} \otimes \dots \otimes a_{n} > O$ for any a₁, a₂, ⋯, a_n ∈ F.

Theorem 3.2 Let F be a proper filter of $L$ . Then F is a maximal filter if and only if it satisfies

(M2)(for any x, y ∈ L) (x ⊕ y ∈ F ⇒ x ∈ F or y ∈ F).

Proof. Assume that F satisfies the condition (M2). Let x ∈ L be such that. Then $x \oplus x^{'} = x^{'} \to x^{'} = I \in F,$ and so x ∈ F by (M2). If, it follows from Proposition 2.2 (F1) that $x \otimes x^{'} \in F,$ that is, O ∈ F which contradicts with F is proper. Hence,. We have F is a maximal filter of $L$ from Theorem 3.1.

Conversely, suppose that F is a maximal filter. Let x, y ∈ L be such that x ⊕ y ∈ F. If x ∉ F, then by (M1). Since F is a filter, it follows from $x^{'} \to y = x \oplus y \in F$ that y ∈ F. Hence, (M2) is valid.

Corollary 3.1. Let F be a proper filter of $L$ . Then F is a maximal filter if and only if it satisfies

(M2)x₁⊕ ⋯ ⊕ x_n ∈ F ⇒ x₁ ∈ F or ⋯ or x_n ∈ F, for any x₁, x₂ ⋯ x_n ∈ L.

Definition 3.3. [21] Let F be a proper filter of $L$ , then F an ultra-filter of $L$ if and only if F satisfies

(M1) x ∈ F if and only if for any x ∈ L.

Remark 3.1. It follows from Theorem 3.1 and Corollary 3.1 that the maximal filter coincide with ultra-filter in lattice implication algebra.

In what follows, let x, a₁, a₂, ⋯, a_n ∈ L and denote

$\underset{n}{\underset{︸}{x \oplus \dots \oplus x}} = nx$ ;

$\underset{n}{\underset{︸}{x \otimes \dots \otimes x}} = x^{n}$ ; $[a_{1}, a_{2}, \dots, a_{n}, x] = a_{1} \to (a_{2} \to (\dots \to (a_{n} \to x) \dots)) .$

Specially,

[a, x] ⁰ = x;

[a, x] ¹ = [a, x] = a→ x;

$[a, x]^{n} = \underset{n}{\underset{︸}{a \to (a \to (\dots \to (a \to x) \dots))}}, (n \geq 2) .$

Proposition 3.1 Let $L$ be a lattice implication algebra, x, y, z ∈ L and m, n ∈ N, then $[x, y \to z]^{m} \otimes [x, y]^{n} \leq [x, z]^{n + m} .$

Theorem 3.3 Let F be a maximal filter of a lattice implication algebra $L$ . Then, for any a ∉ F, there exists n ∈ N such that.

Proof. Assume F is a maximal filter. Define a subset M of $L$ as following: $M = {x | for some y \in F, n \in N, y \leq [a, x]^{n}} .$ Obviously, I ∈ M. If c, c → d ∈ M, then there exist y, h ∈ F, $n, m \in ℕ$ such that y ≤ [a, c] ⁿ and h ≤ [a, c → d] ^m. As F is a filter, we have y ⊗ h ∈ F and $y \otimes h \leq [a, c]^{n} \otimes [a, c \to d]^{m} \leq [a, d]^{n + m},$ therefore d ∈ M, that is, M is a filter of $L$ . For any y ∈ F, y ≤ a → y = [a, y], it follows that y ∈ M, and so F ⊆ M. But I ∈ F and I ≤ [a, a], we also have a ∈ M and F is a proper subset of M. Since F is a maximal filter, this implies M = L. Therefore, O ∈ M. So there exists y ∈ F, n ∈ N such that $y \leq {[a, O]}^{n} = (a^{n})^{'} = n a^{'} .$ As F is a filter and y ∈ F, we have.

Theorem 3.4. Let F be a nonempty subset of $L$ . If F has the finite intersection property, then there exists a maximal filter of $L$ containing F.

Proof. Let M be a maximal filter of $L$ and F₁, F₂ be two filters such that M = F₁ ⋂ F₂. By the maximality of M, we have M = F₁ or M = F₂. It follows that M is a prime filter of $L$ .

Theorem 3.5. Let $L$ be a lattice implication algebra. Then any prime filter of $L$ is contained in a unique maximal filter of $L$ .

Proof. Let the set Φ = {G|G is a proper filter of L containing F }. Obviously, 〈F〉 ∈ Φ, we have Φ is nonempty. Assume that $G_{1} \subseteq G_{2} \subseteq \dots$ is a chain of elements of Φ and let H = ⋃ _iG_i. Therefore, F ⊆ H, O ∉ H and I ∈ H since G_i are proper filter for all i. Let x, y ∈ L such that x ∈ H and x → y ∈ H, then there exists i such that x ∈ G_i and x → y ∈ G_i, it follows that y ∈ G_i since G_i is a proper filter if $L$ . Thus, H is also a proper filter of $L$ containing F and H ∈ Φ. It follows from Zorn’s Lemma that there is maximal element in Φ, this maximal element is the maximal filter of $L$ .

Theorem 3.6. Let $L$ be a lattice implication algebra and F be a proper filter of $L$ . Then F is a prime filter and implicative filter if and only if x ∈ F or for any x ∈ L.

Proof. Suppose x ∈ F or for any x ∈ L. We first to prove F is prime. Let x ∨ y ∈ F. If x ∉ F, then. As and, we have x → y ∈ F. With $x \lor y = (x \to y) \to y \in F$ and F is a filter, it follows that y ∈ F. Therefore, F is a prime filter.

Now we prove F is an implicative filter of $L$ . Let x → (x → y) ∈ F. If x → y ∉ F, we have. As, we have, it follows that x ∈ F. Along with x → (x → y) ∈ F, we have x → y ∈ F, contradiction.Therefore F is an implicative filter of $L$ .

Conversely, assume that F is a prime filter and an implicative filter of $L$ . As $(x^{'} \to x^{'}) \lor (x^{'} \to x) = I \in F$ and F is a prime filter, we have or. If, that is x → (x → O) ∈ F.As F is an implicative filter, we have. If, that is (x → O) → x ∈ F, it follows that x ∈ F.

Theorem 3.7. Let $L$ be a lattice implication algebra and F be a proper filter of $L$ . Then F is a maximal filter and implicative filter if and only if

(M3) x ∉ F and y ∉ F imply x → y ∈ F and y → x ∈ F for any x, y ∈ L.

Proof. Suppose (M3) hold. Assume that F is not an implicative filter, then there exist x, y ∈ L, x ∉ F such that (x → y) → x ∈ F.

(1) If y ∉ F, it follows from condition that x → y ∈ F. As F is a filter, we have x ∈ F, contradiction.

(2) If y ∈ F, as x → y ≥ y, we have x → y ∈ F, it follows that x ∈ F, contradiction.

Now we prove F is a maximal filter. Assume F ⊆ F₁ and F₁ is an implicative filter and there exists x ∈ F₁ such that x ∉ F. Now we need to prove F₁ = L. In fact, for any y ∈ L. If y ∈ F, we have y ∈ F₁; If y ∉ F, it follows from that x → y ∈ F, and so x → y ∈ F₁ and y ∈ F₁. Therefore F₁ = L, that is, F is a maximal filter of $L$ .

Conversely, let x ∉ F and y ∉ F. Since F is an implicative filter, the the set $A_{y} = {t \in L | y \to t \in F}$ is also an implicative filter and A_y is the least filter containing F and y. By the maximality of F, we have A_y = L, therefore x ∈ A_y, that is, x → y ∈ F. Similarly, we can also prove y → x ∈ F.

Corollary 3.2. Let $L$ be a lattice H implication algebra and F be an proper filter of $L$ . Then F is a maximal filter if and only if

(M4) x ∉ F and y ∉ F imply x → y ∈ F and y → x ∈ F for any x, y ∈ L.

3.2 Radical of filters

Definition 3.4. Let F be a proper filter of a lattice implication algebra $L$ . The intersection of all maximal filters of $L$ which contain F is called the radical of F and denoted by Rad(F).

Obviously, if F is maximal filter $L$ , then Rad(F)=F.

Example 3.2. Let $L$ be a lattice implication algebra defined in Example 3.1. F = {I, b, c} is a filter of $L$ and also a maximal filter. According to Definition 3.1, Rad(F)=F = {I, b, c}.

Lemma 3.1. Let F be a filter of a lattice implication algebra $L$ . If a ∈ L \ F, then there is a prime filter P of $L$ such that F ⊆ P and a ∉ P.

Theorem 3.8. Let F be a filter of a lattice implication algebra $L$ . Then $R a d (F) = {x \in L | n x^{'} \to x \in F, for any n \in ℕ} .$

Proof. Let for all n ∈ N and x ∉ Rad (F). Then there exists a maximal filter M of $L$ such that F ⊆ M and x ∉ M. As M is maximal filter of $L$ , there exists n ∈ N such that. We have x ∈ M. Therefore xⁿ ∈ M, furthermore,, i.e., O ∈ M, which is acontradiction.

Conversely, let x ∈ Rad (F) and assume there exists n ∈ N such that. It follows from Theorem 4.1 that there exists a prime filter P of $L$ such that F ⊆ P and. As P is a prime filter of $L$ and $(n x^{'} \to x) \lor (x \to n x^{'}) = I,$ we have. It follows from Theorem 3.4 that there exists a unique maximal filter M of $L$ such that P ⊆ M. Therefore $x \to n x^{'} \in M .$ We can conclude that x ∉ M (In fact, if x ∈ M, then aⁿ ∈ M. In other aspects, as, we have. And thus $x^{n} \otimes n x^{'} \in M,$ that is, O ∈ M, which is a contradiction.). And so F ⊆ P ⊆ M and a ∉ M. therefore a ∉ Rad (F), which is also a contradiction. Therefore, for any n ∈ N.

Theorem 3.9. Let F and G be filters of $L$ and a, b ∈ L. Then the following conditions hold.

(1) If F ⊆ G, then Rad (F) ⊆ Rad (G);

(2) Rad (Rad (F)) = Rad (F);

(3) If a, b ∈ Rad (F), then a ⊕ b ∈ F;

(4) [Rad (F) ∪ Rad (G)) ⊆ Rad (F ∪ G).

Proof. (1) Let F ⊆ G and x ∈ Rad (F), it follows from Theorem 4.2 that $n x^{'} \to x \in F \subseteq G$ for any n ∈ N. Therefore x ∈ Rad (G), that is, (1) holds.

(2) As F ⊆ Rad (F), we have $Rad (F) \subseteq Rad (Rad (F)) .$ Now we need to show $Rad (Rad (F)) \subseteq Rad (F) .$ In fact, for any x ∈ Rad (Rad (F)), Then x ∈ M, where M is a maximal filter of $L$ containing Rad (F). Let M₀ be an arbitrary maximal filter of $L$ containing F, then $M_{0} = Rad (M_{0}) \supseteq Rad (F)$ and so x ∈ M₀. Therefore x ∈ Rad (F), that is, $Rad (Rad (F)) \subseteq Rad (F) .$ And so (2) holds.

(3) Let a, b ∈ Rad (F), then a ⊗ b ∈ Rad (F) and $n (a \otimes b)^{'} \to (a \otimes b) \in F$ for any $n \in ℕ$ . Hence $n (a \otimes b)^{'} \to (a \otimes b) \in F$ As a ⊗ b ≤ a, we have $n (a \otimes b)^{'} \to (a \otimes b) \leq a^{'} \to (a \otimes b) .$ Therefore, As a ⊗ b ≤ b, we have $(a \otimes b) \to b = I \in F,$ and so $a^{'} \to (a \otimes b) \otimes ((a \otimes b) \to b) \in F .$ Since $(a^{'} \to (a \otimes b) \otimes ((a \otimes b) \to b) \leq a^{'} \to b = a \otimes b .$

Thus a ⊕ b ∈ F.

4 Topological properties of maximal filters

In this section, we introduce a new topological space consisted of all maximal filters of a lattice implication algebra $L$ . Let $MF (L)$ be a set of maximal filters of a lattice implication algebra $L$ .

Definition 4.1. Let X ⊆ L. We define a subset $𝕄 (X)$ of $MF (L)$ as follows: $𝕄 (X) = {F \in MF (L) | X ⊈ F} .$

Proposition 4.1. For any X, Y ⊆ L and {X_i} _i∈I be some subsets of L, then the following propertieshold:

(1) If X =∅ or X = {I}, then $𝕄 (X) = \emptyset$ ;

(2) If X ⊆ Y, then $𝕄 (X) \subseteq 𝕄 (Y)$ ;

(3) $\cup_{i \in I} 𝕄 (X_{i}) = 𝕄 (\cup_{i \in I} X_{i})$ ;

(4) $𝕄 (X) = MF (L)$ if and only if [X) = L.

Proof. (1) For any filter F of $L$ , we have I ∈ F and ∅ ⊆ F. It follows that $𝕄 ({I}) = 𝕄 (\emptyset) = \emptyset .$ Therefore, $𝕄 (X) = \emptyset$ when X =∅ or X = {I}.

(2) The proof of (2) is trivial.

(3) It follows from (2) that $\cup_{i \in I} 𝕄 (X_{i}) \subseteq 𝕄 (\cup_{i \in I} X_{i}) .$ For any $F \in 𝕄 (\cup_{i \in I} X_{i})$ , there exist i₀ ∈ I such that X_{i
₀} ⊈ F, that is $F \in 𝕄 (X_{i_{0}})$ , therefore $F \in \cup_{i \in I} 𝕄 (X_{i})$ . Hence $𝕄 (\cup_{i \in I} X_{i}) \subseteq \cup_{i \in I} 𝕄 (X_{i}) .$ Thus $\cup_{i \in I} 𝕄 (X_{i}) \subseteq 𝕄 (\cup_{i \in I} X_{i})$ .

(4) Suppose that $𝕄 (X) = MF (L)$ . If [X) ≠ L, then there exist a maximal filter F such that X ⊆ [X) ⊆ F, then $F \notin 𝕄 (X)$ , which contradicts with $𝕄 (X) = MF (L)$ . Therefore [X) = L.

Conversely, for any $F \in MF (L)$ , since [X) is a minimal filter containing X, we have X ⊆ F if and only if [X) ⊆ F, that is, X ⊈ F if and only if [X) ⊈ F. Therefore $𝕄 (X) = 𝕄 ([X))$ . When [X) = L, we have $𝕄 (X) = 𝕄 ([X)) = 𝕄 (L) = MF (L) .$

When X = {x|x ∈ L}, we denote $𝕄 ({x})$ by $𝕄 (x)$ , that is $𝕄 (x) = {F \in MF (L) | x \notin F} .$

Proposition 4.2. For any x, y ∈ L, $𝕄 (x)$ satisfies the following properties:

(1) $𝕄 (I) = \emptyset$ ;

(2) $𝕄 (O) = MF (L)$ ;

(3) If x ≤ y, then $𝕄 (y) \subseteq 𝕄 (x)$ ;

(4) $𝕄 (x) \cap 𝕄 (y) = 𝕄 (x \lor y)$ ;

(5) $𝕄 (x) \cup 𝕄 (y) = 𝕄 (x \land y) = 𝕄 (x \otimes y)$ .

Proof. (1) and (2) are obvious, the details are omitted.

(3) Let x, y ∈ L and x ≤ y. For any $F \in 𝕄 (y)$ , then y ∉ F. As F is a filter of $F$ and x ≤ y, it follows that x ∉ F. Therefore $F \in 𝕄 (x)$ , that is, (3) hold.

(4) For any x, y ∈ L, on the one hand, it follows from (3) that $𝕄 (x \lor y) \subseteq 𝕄 (x) \cap 𝕄 (y) .$ On the other hand, for any $F \in 𝕄 (x) \cap 𝕄 (y)$ , then x ∉ F and y ∉ F. Since F is a maximal filter, we have F is a prime filter. It follows that x ∨ y ∉ F, that is, $F \in 𝕄 (x \lor y)$ . Therefore $𝕄 (x) \cap 𝕄 (y) \subseteq 𝕄 (x \lor y)$ and $𝕄 (x) \cap 𝕄 (y) = 𝕄 (x \lor y)$ .

(5) For any $F \in MF (L)$ and x, y ∈ L, we have x ⊗ y ∉ F if and only if x ∧ y ∉ F if and only if x ∉ F or y ∉ F. In fact, let x ⊗ y ∉ F, if x ∧ y ∈ F, it follows from F is a filter that x ∈ F and y ∈ F, therefore x ⊗ y ∈ F, which contradicts with x ⊗ y ∉ F. On the contrary, if x ∧ y ∉ F, then x ⊗ y ∉ F for x ⊗ y ≤ x ∧ y and F is a filter of $L$ . Therefore $𝕄 (x \otimes y) = 𝕄 (x \land y) = 𝕄 (x) \cup 𝕄 (y) .$ So (5) hold.

Now we define a set B as follows:

$B = {𝕄 (x) | x \in L}$ .

Proposition 4.3. B is a base of some topology $T$ .

Proof. It follows from Proposition 4.2 that $𝕄 (O) = MF (L) \in B$ , and so $\cup B = MF (L)$ . For any $𝕄 (x), 𝕄 (y) \in B$ and $F \in 𝕄 (x) \cap 𝕄 (y)$ , then there exist $𝕄 (x \lor y) \in 𝔹$ such that $F \in 𝕄 (x \lor y) \subseteq 𝕄 (x) \cap 𝕄 (y) .$ Therefore, B is a base of some topology $T$ .

Theorem 4.1. $T = {𝕄 (X) | X \subseteq L}$ .

Proof. For any X ⊆ L, it follows from Proposition 4.1(3) that $𝕄 (X) = 𝕄 (\cup_{x \in X} {x}) = \cup_{x \in X} 𝕄 (x)$ . By Proposition 4.1(1), we have $𝕄 (X)$ is a open set of $T$ .

Conversely, let $U$ be a open set of $T$ , there exist $B_{U} \subseteq B$ such that $\begin{matrix} U & = \cup_{𝕄 (x) \in B_{U}} 𝕄 (x) \\ = 𝕄 (\cup_{𝕄 (x) \in B_{U}} {x}) \\ \in {𝕄 (X) | X \subseteq L} . \end{matrix}$ Hence, $T = {𝕄 (X) | X \subseteq L}$ .

Theorem 4.2. ( $MF (L), T$ ) is a compact topology space.

Proof. Let $C$ is an open cover of $MF (L)$ and $C \subseteq T$ . As B is a base of $T$ , then there exist ${𝕄 (x_{i}) | i \in I} \subseteq B$ such that $MF (L) = \cup C = \cup_{i \in I} 𝕄 (x_{i}) .$ Therefore, $MF (L) = \cup_{i \in I} 𝕄 (x_{i}) = 𝕄 ([\cup_{i \in I} x_{i})) .$ It follows that [∪ _i∈Ix_i) = L, hence O ∈ [∪ _i∈Ix_i). this means that there exist $a_{i_{1}}, \dots, a_{i_{n}} \in \cup_{i \in I} {x_{i}}$ such that a_{i
₁} ⊗ ⋯ ⊗ a_{i
_n} = O. Therefore $\begin{matrix} MF (L) & = 𝕄 (O) \\ = 𝕄 (a_{i_{1}} \otimes \dots \otimes a_{i_{n}}) \\ = 𝕄 (a_{i_{1}}) \cup \dots 𝕄 (a_{i_{n}}) . \end{matrix}$ Hence there exist a finite subcover set {C_{i
₁}, ⋯, C_{i
_n}} such that $𝕄 (x_{i_{1}}) \subseteq C_{i_{1}}, \dots, 𝕄 (x_{i_{n}}) \subseteq C_{i_{n}},$ and so ( $MF (L), T$ ) is a compact topology space.

Theorem 4.3. ( $MF (L), T$ ) is a Hausdorff space.

Proof. Let F₁, F₂ be any two maximal filters of $L$ and F₁ ≠ F₂, there exist x, y ∈ L such that $x \in F_{1} - F_{2}, y \in F_{2} - F_{1} .$ We have $x \to y \notin F_{1}, y \to x \notin F_{2}$ (otherwise, y ∈ F₁, x ∈ F₂ which contradict with y ∉ F₁, x ∉ F₂). Therefore, $\begin{matrix} F_{1} \in 𝕄 (x \to y) \in T, \\ F_{2} \in 𝕄 (y \to x) \in T, \end{matrix}$ that is, $𝕄 (x \to y)$ and $𝕄 (y \to x)$ are the open neighborhood of F₁ and F₂, respectively. And so $\begin{matrix} 𝕄 (x \to y) \cap 𝕄 (y \to x) & = 𝕄 ((x \to y) \lor (y \to x)) \\ = 𝕄 (I) = \emptyset . \end{matrix}$

Then ( $MF (L), T$ ) is a Hausdorff space.

5 Conclusions

In order to research the many-valued logical system whose propositional value is given in a lattice, Xu initiated the concept of lattice implication algebras. Hence for development of this many-valued logical system, it is needed to make clear the structure of lattice implication algebras. It is well known that to investigate the structure of an algebraic system, the filters with special properties play an important role. In this paper, some equivalent characterizations of maximal filters are given, firstly. Finally, by introducing some topological structures on the set of all maximal filters and by investigating the topological properties of them, we conclude that the set of all maximal filters is a compact topological space and the set of all maximal filters is a Hausdorff topological space. We hope that above work would serve as a foundation for enriching corresponding many-valued logical system.

Future work plans include theoretical extensions driven by practical applications. In particular, on the one hand, regarding theory, extensions of this work might also be sought in the mathematical lattice of fuzzy numbers based on their (interval) a-cuts; likewise extensions can be sought in the mathematical lattice of Intervals Numbers. On the other hand, regarding practical applications, machine intelligence might be pursued, e.g. in mobile/humanoid robot applications, toward planning by reasoning that accommodates uncertainty.

Footnotes

Acknowledgments

This work is supported by National Natural Science Foundation of P.R.China (Grant no. 61673320); The Application Basic Research Plan Project of Sichuan Province (No.2015JY0120); The Scientific Research Project of Department of Education of Sichuan Province (15TD0027, 18ZA0273, 17ZB0220); Natural Science Foundation of Guangdong Province (2016A030310003).

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